# How to simplify this summation expression

The expression is:

$$\sum_{k=1}^{n} 1 - \Bigg(\sum_{k=1}^{n} 2^{-k/2}\Bigg)^2$$

I know that $$\sum_{k=1}^{n} 1 = 1n$$

I don't really know how to proceed with the latter part.

• You have a geometric series – Jacky Chong Oct 19 '16 at 20:12

$$\sum_{k=1}^{n} 2^{-k/2}=\sum_{k=1}^{n} \left(\frac{1}{\sqrt{2}}\right)^k,$$ which is a geometric series (see Eq. 8). Can you take it from here? If not, please do let me know.
• @RyanSmith Please don't just repeat an already existing question. From what you write ($k=1,2,3,\dots$), it seems $k$ runs from $1$ to $\infty$, but this also means that your variance is infinite. – Bobson Dugnutt Oct 20 '16 at 0:04